Download Hitting 10

Hitting 10
You toss a coin repeatedly. If heads is worth 1 and tails is worth 2, and
you add up the score as you go along, what is the probability that you’ll
hit 10?
That is to say, 10 might be one of the sums you get, but it also might
not—if you hit 9 and then roll tails, you’ll go straight to 11 and miss
10. The question asks, if you do this a large number of times, what
proportion of the times will you get 10 as one of your sums?
A note on notation: it makes sense to jettison the terminology “heads”
and “tails” and use “1” and “2”, as if the coin had a 1 on one side and a
2 on the other. This is what I will do.
As usual, I start out collecting data. Each of 35 students runs 10 trials
and keeps track of the number of “successes” (hitting 10). The scores
range from 4/10 to 9/10 successes, with an overall success rate of
216/350 = 0.62 . That’s an empirical estimate of our probability.
One student asserts right away that the answer should be 2/3. Here is
her argument. Think of the possible sequences of sums that go “past”
10. We could have
8, 10
9, 10
9, 11
These are the only possibilities—that is, every sequence will contain
one (and only one) of these pairs. Now two of these hit 10 and one
misses 10, for a probability of 2/3. An interesting argument.
But a number of students are uncomfortable with it and there is a clear
need for some care in the analysis. But where to begin? Someone
suggests that if we knew the probability of hitting 8 or 9, then we might
be able to figure out the probability of hitting 10. That sounds like
induction—working up from small numbers. So let’s take some small
numbers N and see what we get.
What’s wrong with this argument? That’s not easy to
see at first. In fact probabilistic arguments can be devilishly tricky. It is true that a
probability can be regarded
as the proportion of “successes” among all the “possibilities” but it’s important
that these possibilities all be
equally likely. Are these
three “sequences” equally
likely to occur?
Some notation: we’ll denote by PN the probability of hitting the target N.
Why not start at the very beginning. What’s the probability of hitting
N=1? Well, to hit 1, you’ll have to toss 1 right away, and that will
happen with probability ½. So P1 = ½ .
Okay, let’s go for N=2. Here I get a verbal argument. Think about the
first toss: half the time you toss 2 and succeed right away. What about
the other half when you toss 1? Well, then you toss again and half the
time you toss another 1 (success) and the other half you toss 2 (failure).
So with probability ½ + ¼ you hit 2 and with probability ¼ you miss it.
So P2 = ¾.
hitting10
11/13/2007
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Now on to N=3. This time a student comes to the board and tabulates
all the possible sequences that would arise, stopping each time when it
is known whether 3 has been hit. There are 5 possible sequences, of
which 3 are successful, for a probability of 3/5. I let the table stand,
and take a vote. Nearly everyone accepts the result. But there are a
few abstainers with cross looks and wrinkled brows.
Buoyed up by this show of support, the student carries on. He has
done the same for N=4 and found 8 possible sequences of which 5 are
successes. Now 3, 5 and 8 are successive Fibonacci numbers, and he
asserts that it continues in this way. Thus PN is always the ratio of two
successive Fibonacci numbers.
The N=3 table
111
S
112
F
12
S
21
S
22
F
Well, what can I say? Wow. Certainly the class is cowed. “Comments?” I ask crisply. Silence, even from the wrinkled brows. We’ve
encountered the Fibonacci numbers. What could be better? I just has
to be right.
The N=4 table
1111
S
1112
F
112
S
121
S
122
F
211
S
212
F
22
S
To trouble-shoot, I suggest we go back and do the same analysis for the
case N=2 that we have already done. It shows three possibilities, two
of which are successes. Thus P2 should be 2/3. But we’ve just shown
that P2 = 3/4. So there’s certainly a flaw in the reasoning here.
The N=2 table
11
S
12
F
2
S
Let’s compare this with the original argument—there we noted that the
initial 2 (the third row) occurs ½ the time, so this contributes ½ to the
probability right away. And the other two cases each contribute ¼.
That is, it would seem that the rows of the table should be weighted
with their probability of occurrence. If we include these weights (as a
third column), and add up all the S-weights, we do get the correct answer of ¾.
The N=2 table with
weights
11
S
¼
12
F
¼
2
S
½
It would appear that the flaw in the “Fibonacci” calculation is the various rows in the table haven’t been weighted by their probability of occurrence. Let’s see what happens if we do that with the N=3 table.
The new weighted table now gives a success probability of 1/8 + 1/4 +
1/4 = 5/8 and this is in fact the correct answer. It also checks out
with a couple of approaches that other students have been trying.
Note that since the third column represents the probabilities of the different possible outcomes, they have to sum to 1. That’s a nice error
check. [We’ve had to abandon that amusing Fibonacci excursion, but
see problem 3.]
hitting10
11/13/2007
The N=3 table with
weights
111
S
1/8
112
F
1/8
12
S
1/4
21
S
1/4
22
F
1/4
2
1 S
F
2
1
1
1
I give the class the job of constructing the weighted table for N=4, and
we get the answer of 11/16. It starts to get hard to organize the entries
in a systematic way. Some students are using tree diagrams instead of
tables, and they provide an easy system, but they get large quickly.
S
2
1
1
S
2
1
The N=4 table
with weights
1111
S
1/16
1112
F
1/16
112
S
1/8
121
S
1/8
122
F
1/8
211
S
1/8
212
F
1/8
22
S
1/4
2
2
1
2
F
2
1
1
2
1
2
S
F
2
1
2
1
Can you track the correspondence
between the tree diagram and the table?
1
2
S
2
1
2
2
Perhaps it’s time to start tabulating the PN values to see if we can see
some useful patterns in the numbers. So far, we have calculated the
first 4 values. Any guesses as to what comes next? One clear pattern
is seen in the denominators—2, 4, 8, 16––at the Nth stage we get 2N.
Well, an interesting pattern is observed in the numerators as well: 1, 3,
5 and 11. Add consecutive pairs:
1+3=4
3+5=8
5 + 11 = 16
N
1
2
3
4
5
PN
1/2
3/4
5/8
11/16
?
N
1
2
3
4
5
6
7
8
9
10
PN
½
¾
5/8
11/16
21/32
43/64
85/128
171/256
341/512
683/1024
which is the denominator of 3
which is the denominator of 5
which is the denominator of 11.
Can this be a coincidence? The prediction it gives for P5 is 21/32.
And someone has been constructing the table for N=5, and reports that
this is correct! Wow. Nice.
Perhaps we should stop now and try to see why this interesting numerical pattern might hold (and thereby get a proof of its validity). But the
class is impatient to complete the table—and so that’s what we do.
Okay! We have a candidate: P10. = 683/1024. How can we be sure
that it’s correct?
A hand goes up in the back. It is Devon.
It’s correct.
I beg your pardon?
It’s correct. The answer for P10 is correct.
How do you know?
I worked it out.
hitting10
11/13/2007
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You worked it out? You calculated P10?
Yes.
How?
Uhmm…
You made a table of all the possibilities?
Not exactly…
A tree?
Not really…
I invite Devon to the front, and give him the chalk and a clean panel.
He proceeds to display an extremely elegant accounting of all possible
paths leading to a success for N=10. With a bit of reorganization on
my part, this is the table he puts up.
sequence
composition
10 ones
8 ones & 1 two
6 ones & 2 twos
4 ones & 3 twos
2 ones & 4 twos
5 twos
Different routes to get to 10
Prob. of
number of
total
sequence
sequences
probability
½10
1
1/210
9
½
9
9/29
8
½
8!/6!2! = 28
28/28
7
½
7!/4!3! = 35
35/27
6
½
6!/2!4! = 15
15/26
5
½
1
1/25
Having put the table on the board, he then calculates the sum:
1
10
2
+
9
9
2
+
28
8
2
+
35
2
7
+
15
2
6
+
1
25
1
683
= 10 [1 + 2 ⋅ 9 + 4 ⋅ 28 + 8 ⋅ 35 + 16 ⋅ 15 + 32] = 10
2
2
A convincing result!
Needless to say, this “full calculation” is not the route I intended the
class to follow. Indeed, many of them have not yet encountered the
combinatorial coefficients. And I can feel a definite level of anxiety
rising from the class.
What happened here really
blew me away. Devon hadn’t
wasted any time either—he had
in fact obtained the answer
before we had completed the
table, and was just waiting for
the chance to put it forward. It
turns out he was taking OAC
finite math that very term (the
year was 1995), and this problem hit him right at a golden
moment.
Most of the class has trouble
seeing what Devon has done.
It helps to look back at the N=4
table. Notice, for example, that
there are three rows which have
2 ones and 1 two, each with
probability 1/8. Well, in his
accounting, these would all be
listed on one row and there
would be a 3 in his third column to say that he’s counted
three rows of the “expanded”
table. For example, the 28 in
column 3 says that there are 28
ways to arrange a string of 6
ones and 2 twos, and so these
would have occupied 28 rows
in the extended table. So the
table at the left gives us a very
economical presentation.
“What we have here,” I say, “Is a complete (and remarkably compact)
listing of cases.” Such an approach is of great theoretical importance,
and you would encounter it in any introductory finite math course. So
I’m very happy for you to have seen an example here of what it looks
like. However the method has some disadvantages. It is computationally heavy. How well would it work if I had asked for the probability
of hitting 100?” I look at Devon and he wrinkles his brow. “And even
more––could this approach give us a general formula for PN?”
In fact in problems of this type there is often an approach which does
not require listing all the successful cases, which uses a nice piece of
reasoning to get around that. Let’s get back to some of those intriguing
patterns.
hitting10
11/13/2007
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By now, the class has noticed more patterns in the PN table. For example, if you double any numerator, you get the next numerator ±1,
where you alternate the + and the – .
The point is that these numerical patterns should correspond to algebraic relationships between the PN.. And if we can formulate these, we
might be able to “see” why they must hold.
N
1
2
3
4
5
PN
1/2
3/4
5/8
11/16
21/32
Let’s take, for example, the original numerator pattern:
1+3=4
3+5=8
5 + 11 = 16, etc.
What does this tell us about relationships among the PN ? To have a
specific case, let’s take:
5 + 11 = 16.
To get the PN into the picture, divide by 16:
5 11
+
=1
16 16
Now the 11/16 is clearly P4 . But what is the 5/16. Well, it would
seem to be half of P3. So we write this as:
P3
+ P4 = 1
2
Okay. That’s an intriguing relationship. Who can see a reason why it
should be true?
Well, since the two terms on the left sum to 1, and the second is the
probability of hitting 4, the first has to be the probability of not hitting
4. So that becomes the question—can we see a reason why P3/2 should
be the probability of not hitting 4?
Well, it takes a minute, but it’s all too easy. The crucial observation is
that there’s only one way to miss 4 and that’s to hit 3 and then toss a 2.
And the chances of doing that are P3 × ½, the first to hit the 3, and the
second to toss the 2. That’s real nice.
This is a general result: The only way to miss n+1 is to hit n and then
toss a 2. This can be written:
1 − PN +1 =
1
PN
2
PN +1 = 1 −
1
PN
2
If we rewrite it as:
It is in recursive form and this allows us to recursively calculate all the
PN starting at P1 = ½.
hitting10
11/13/2007
I give the students some
time to find this lovely argument on their own, but
not many succeed. They
seem to find it a bit foreign.
One might have thought that
this recursive approach would
have been tried at the very
beginning of the problem. But
for all it’s simplicity and directness, it does not seem to
occur naturally. It’s an approach that must be learned
and experienced.
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One more question––can we work out P100 from this? Of course we
could start with P1 and iterate the recursion 99 times. That would be
easy enough with a spreadsheet, but where would it get us? What we’d
really like is a formula for PN in terms of N. And that’s really a question of solving the recursion.
We have what’s called a first-order linear inhomogeneous recursion
and there are lots of different “methods” around to solve it. But my
favorite is the one that I find high school students doing if they’re not
told what to do, and that’s just to iterate the recursion again and again
but keeping the terms in expanded form until one sees a pattern.
1
PN
2
We have a recursive formula
for PN. Can we solve it?
PN +1 = 1 −
To make the calculations easier (and more transparent!) we write the
recursion as
PN+1 = 1 + aPN
with a = – ½ .
Then start at the beginning:
Notice what a difference that “a”
makes. Using a general parameter
there instead of the –½ really allows us to focus on the developing pattern as we move from one
N to the next.
P0 = 1
P1 = 1 + aP0 = 1 + a.
P2 = 1 + aP1 = 1 + a(1 + a) = 1 + a + a2
P3 = 1 + aP2 = 1 + a(1 + a + a2) = 1 + a + a2 + a3
P4 = 1 + aP3 = 1 + a(1 + a + a2 + a3) = 1 + a + a2 + a3 + a4
The pattern is clear and it is clear that it will continue (a formal proof
of that can be made using mathematical induction).
PN = 1 + a + a2 + a3 + ... + aN =
1 − a N +1
1− a
using the formula for the sum of a geometric series.
Now putting a = – ½:
PN =
[
]
1 − (−1 / 2) N +1
1 − (−1 / 2) N +1
2
=
=
1 − (−1 / 2) N +1 .
3
3/ 2
1 − (−1 / 2)
We have a formula! The probability of hitting 10 is then
P10 =
[
2
1 − (−1 / 2)11
3
]
=
683
2⎡
−1 ⎤
2 ⎡ 2049 ⎤
1−
=
=
.
⎢
⎥
⎢
⎥
1024
3 ⎣ 2048 ⎦
3 ⎣ 2048 ⎦
as we obtained with our combinatorial calculations.
The general formula is nice. It tells us that the probabilities will be
successively above and below 2/3, and will approach 2/3 in the limit.
You might have managed to already argue that the probability should
get close to 2/3 for large N.
Why should the PN
approach 2/3?
Here’s a nice way of thinking
about it. Suppose we make a
large number of tosses, say 200.
Then we expect 100 1’s and 100
2’s, so our total sum should be
about (on average?) 300. But
how many numbers between 1
and 300 will we have hit?––well
we get a new hit with every toss,
so we’ll hit 200 numbers. That
means we hit a proportion
200/300 of the numbers less than
300. So the probability of hitting
any particular number should be
around 2/3.
Question. There’s something a bit unexpected about the PN formula
and that’s the 3 in the denominator. Isn’t the denominator supposed to
be 2N ? Argue that PN will always be able to be written as a fraction
with 2N in the denominator.
hitting10
11/13/2007
6
Problems
1. Harry and Sally play the following game with a biased coin. Sally flips first. If she gets heads she
wins; if she gets tails she gives the coin to Harry. Then Harry flips and if he gets heads he wins; if he gets
tails he gives the coin to Sally. The game continues until someone flips heads and wins.
(a) If heads comes up with probability 1/4, what is the probability Sally will win?
(b) If over many repetitions of the game, Sally wins 75% of the time, what would you say was the bias on
the coin?
(c)Let the bias on the coin be such that heads occurs a proportion x of the time, and suppose Sally wins
with probability p. Find simple formulas for p in terms of x and for x in terms of p.
2. Annie and Bill each have a fair die, which they throw alternately. Annie scores if she throws a multiple
of three, and Bill scores if he throws a multiple of two. The winner is the first person to score. Show that
the game is fair if Annie rolls first.
[Note: If I'm using this problem in class, it works well to put the class into groups of two to play this game
a few times, and collect some data. We see if the game really seems to be fair, and we get a "hands-on"
feel for the analysis.]
3. Here’s a generalization of 2. Suppose that Annie and Bill each spin a pointer with n equally likely outcomes numbered 1 to n. Annie scores if she throws any of the first numbers 1, 2, … k and Bill scores if he
throws any of the numbers k+1, k+2, … n. If Annie goes first, for what numbers n and k is the game fair?
4. A flips a fair coin 100 times, and B flips a fair coin 101 times. What is the probability that B gets more
heads than A? [This is an awesome problem and works well with an exploratory approach. Start by solving simpler problems.]
5. Following Fibonacci. What the Fibonacci guy discovered (above) is that if you count the number of
different sequences that you might generate in deciding whether a target n was “hit” you got a Fibonacci
number, and if you counted the number of those that were “successful,” in that they hit the target, that was
the preceding Fibonacci number. In fact, that’s a very nice result, not about probabilities, but about numbers of sequences of a certain type. Track these ideas down and write a nice little essay about the subject.
6. Follow the methods developed above to solve the recursion
x n +1 = 3 −
3
x
4 n
starting with x0 = 1. That is, find a general formula for xn .
hitting10
11/13/2007
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